Problem: Integrate. $\int\left(-3e^x+\dfrac4x \right)dx=\,?$ Choose 1 answer: Choose 1 answer: (Choice A) A $-3e^x+4x+C$ (Choice B) B $-3e^x+4\ln|x|+C$ (Choice C) C $-3\ln|x|+4\ln|x|+C$ (Choice D) D $-3\ln|x|+4x+C$
Solution: We can integrate using the following formulas for the indefinite integrals of $e^x$ and $\dfrac1x$ : $\begin{aligned} &\int e^x\,dx=e^x+C \\\\ &\int \dfrac1x\,dx=\ln|x|+C \end{aligned}$ $\begin{aligned} &\phantom{=}\int\left(-3e^x+\dfrac4x \right)dx \\\\ &=-3\int e^x\,dx+4\int\dfrac1x \,dx \\\\ &=-3e^x+4\ln|x|+C \end{aligned}$